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How To Connect Two PSUs, To Get More Current - Solved
Posted by
t1d
on 10 Nov, 2018 22:40
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In parallel, I would think. I need more current, to test an electronic load design.
Are there any safety concerns?
Precision of matching the voltages?
Floating grounds, etc? The DUT is floating.
Any tips are appreciated.
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#1 Reply
Posted by
james_s
on 11 Nov, 2018 00:59
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What kind of power supplies?
Speaking generally, it's one of those things you can often get away with doing, but not something that is recommended. If you want to try it, just beware that you may damage one or more of the power supplies, or it may work just fine, can't say for sure without trying.
There are of course power supplies specifically designed for parallel load sharing, if 12V will do then most any hot swap server PSU has that capability and they are cheap as muck on the surplus market. They're quite powerful too, 600W-1kW is common.
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#2 Reply
Posted by
Audioguru
on 11 Nov, 2018 01:58
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If they have voltage regulation then if one detects that the voltage is too high then it shuts down until the overloaded other regulator drops the voltage enough for this one to turn on. Then they might shuffle the load back and forth without increasing the current.
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#3 Reply
Posted by
t1d
on 11 Nov, 2018 05:14
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Thank you, both, for your replies.
Doh, good question, James. I should have thought to include that information... They are both lab-type power supplies. Tenma and BK Precision...
They are just basic models... Meaning they can set either voltage or current to constant mode, but not both, at the same time. Both are Linear/transformer and 30v/3a.
As Audioguru says, for whichever is set to constant, the other is varied, to maintain the constant. For the test that I need to do, I do not want either to vary. The way I get that to happen is to set the current to constant, above the desired current level test point that I will drive the electronic load's current draw to. So, current and voltage are both maintained, unless I draw too much current, with the e-load, which I do not do, as I am watching the meters and the test level is below the constant current setting.
The problem is that PSUs say 30v/3a, but that is not simultaneously. The available wattage is the number of importance. I need more wattage, than I have available. That is why I am considering connecting them.
If I needed more voltage, I would just connect them in series. I know how to do that safely. But, I have never needed more current and, therefore, I have never connected them in parallel.
James, what kind of problems might come up? How can I avoid those problems?
Thanks, guys.
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if they are current limited psu, you can set both to exactly same voltage, connect them together and watch the current draw from each psu. if they can share load nicely, then you can do it. i think some precision psu is push pull type meaning when there is overvoltage, they will try to stabilize the voltage by pulling excess current to their ground, so two psu with different voltage setting, the one will try to keep up the voltage while the other will sink the current hard to the ground. but if stabilization is done only by a simple resistor, then it will be much less disaster. in case you want to be safe and/or the psu is not current limited type, no display to see current draw etc, put beef of schottky diodes rated for the intended current draw at each the psu output. say if you want to draw 20A from both psu combined, and you have bunch of 1A diodes stock, then try putting 10 diodes on each psu output, more is better for safety margin. ymmv.
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#5 Reply
Posted by
Nusa
on 11 Nov, 2018 06:26
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Speaking of things worth mentioning, what are your load requirements?
You may be able to get there with series instead of parallel, without as much risk of it not working.
(15V+15V)@2A is the same result as 30V@(1A+1A). Especially since you implied the limit was the overall wattage of the supplies, not their maximum settings.
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#6 Reply
Posted by
t1d
on 11 Nov, 2018 06:29
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Excellent, Mechatrommer! Problem solved.
I found a good video, that covers your points.
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#7 Reply
Posted by
t1d
on 11 Nov, 2018 06:36
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Speaking of things worth mentioning, what are your load requirements?
You may be able to get there with series instead of parallel, without as much risk of it not working.
(15V+15V)@2A is the same result as 30V@(1A+1A). Especially since you implied the limit was the overall wattage of the supplies, not their maximum settings.
I am testing a constant current electronic load design. For the Single-MOSFET test board, I want 30v/2a = 60w. For the Two-MOSFET final design, I want 30v/4a = 120w. But, these limits are a bit of design goal guess work. Testing will determine the actual SOA.
But, even with my two PSUs in parallel, I want get there, because they just don't have enough watts.
Thanks for your post.
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I found a good video, that covers your points.
be aware that isolating psu's with resistors or diodes will have voltage drop on the load side, so you have to compensate (increase psu output voltage) and measure manually voltage applied to the load as psu's will have no way of compensating this unless you have psu's with sense input, but if you have many psu's with sense connecting together then things can get complicated quickly.
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#9 Reply
Posted by
t1d
on 11 Nov, 2018 07:09
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Good points. Thanks.
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#10 Reply
Posted by
mvs
on 11 Nov, 2018 09:26
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They are just basic models... Meaning they can set either voltage or current to constant mode, but not both, at the same time. Both are Linear/transformer and 30v/3a.
There is no power supply available, that can control the resistance of an arbitrary load. So its fine to have either CC or CV, but not both at the same time
Set desired voltage on both power supplies using some load and same DMM, then connect them in parallel. The resistance of connecting wires will do the load balancing.
PS It applies to lab power supplies that can only source current. If your PSUs have Push-Pull output, add diods or disable the current sink.
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#11 Reply
Posted by
t1d
on 13 Nov, 2018 01:23
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Okay, I have been studying on this. I understand the use of resistors... Diodes? Not so much...
If using diodes, I do understand using Schottky's... They are faster... But, I am not sure I have any, on-hand.
I have also seen YT video tutorials that say "blocking" diodes can be used. I think this is just a term, in the solar power community, for bigger common rectifying diodes. Do I have that correctly? I do have 1N4004s and 1n4007s, in stock. And, maybe, 1N4002s...
I look forward to trying this, with diodes. I just have to get comfortable, with the idea... I don't want to hurt the new circuit that I have worked so hard on and am so proud of, my (precious; expense vs. personal financial budget) power supply, or me... lol
Thanks, for your continued help.
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#12 Reply
Posted by
t1d
on 13 Nov, 2018 16:22
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I did try 1N4007 rectifying diodes. I could not get those to balance the two PSUs. So, I am looking at buying various parts/options; Schottky Diodes and Resistors.
With the supply only partially balanced, I could tell that the electronic load DUT can handle more current. So, woot, to that.
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#13 Reply
Posted by
macboy
on 14 Nov, 2018 14:10
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You don't need to balance the two supplies if they have proper current limiting, which means that they will automatically go into constant current mode when overloaded. Essentially, you will allow one supply (whichever one tends to supply more current) to go into constant current mode and the other will then 'take over'. See below.
If you are using a basic dumb SMPS which has only over-current protection (like a computer power supply), this won't work. When the first supply overloads, it will shut down, then the other one will become overloaded and shut down. In that case, you need to very carefully balance the load, by matching output voltages and using balance resistors. The easiest type of high current low-ohm resistor is a sizable length of suitable copper wire. Actually, two lengths of wire: one between each supply and the load.
You can also use the copper wire balancing resistor trick to help balance the load between your two lab supplies, but you don't really need to. When you apply an increasing load (which means a decreasing load resistance) to two parallel supplies, the supply with a slightly higher voltage will supply all/most of the current until it reaches its current limit. Then its output voltage will begin to drop (which is exactly what must happen in constant current mode when load resistance continues to drop, because V=I*R). When that voltage reaches the slightly lower voltage of the other supply, that supply will begin to source current into the load, keeping the voltage steady at that point. If the load continues to increase (resistance decreases), then the second supply will also eventually reach its current limit, and the voltage will begin to drop again. You can see this automatic crossover at work very clearly if you set up one supply to a clearly higher voltage than the other, say 10 V and 12 V. Experiment with that setup to convince yourself that the two supplies can work in parallel without issue. Then set up both to the same(ish) voltage and let them take care of sharing the load. One will switch to constant current mode before the other, but that is OK.
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#14 Reply
Posted by
t1d
on 15 Nov, 2018 00:12
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MacBoy, that is a brilliant explanation! Thank you.
I have a couple of Schotty's on order, at my local supplier. They will give me more options, in addition to the copper wire trick.
The PSUs are both linear. You said that one of the PSUs will follow the other. This coincides, with my original test observations.
Let's talk wattage. The PSUs are only capable of so many watts. As you explained, at their max watts, any change in current is offset by a change in voltage and visa versa. So, at the moment the PSUs have gone into constant current mode, this is the combined max wattage output. Therefore, this is the point where the e-load/DUT will see everything the power supply combo has to offer. There is no need to push beyond this point, because I would only be changing the v/c ratio. Correct? The e-load is CC, too.
One PSU has the constant current function. The other has both constant current and constant voltage. Is there any trick to combining these somehow, to an advantage, for this type testing?
Eventually, I think I will need more watts, than what the two PSUs can supply. The e-load design goal is 30v @ 3a. Buck converters are cheap and can supply a lot more. Is there any reason to not use one? They are switch mode, but it would be on the load side, of the circuit...
Thank you, so much, for your excellent help.
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schottky diode has low voltage drop, so it can handle more current without burning away (diode power dissipation) you can use any normal diode it just it will get more hotter than schottky at high current, ymmv.
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#16 Reply
Posted by
t1d
on 15 Nov, 2018 10:53
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schottky diode has low voltage drop, so it can handle more current without burning away (diode power dissipation) you can use any normal diode it just it will get more hotter than schottky at high current, ymmv.
Yes, the 1N4007s did get very hot.